The initial rate for a first order reaction is $0.6932 \times 10^{-2} \ mol \ L^{-1} \ min^{-1}$ and the initial concentration of the reactant is $0.1 \ M$. Then $t_{1/2}$ is equal to ...... $min$.

The integrated rate equation is $Rt = \log C_0 - \log C_t$. The straight line graph is obtained by plotting

$A$ reaction has a rate constant $k = 2.4 \times 10^{-4} \ s^{-1}$. Find the ratio of $t_{99.9}$ to $t_{50}$.

Decomposition of $H_2O_2$ follows a first order reaction. In $50 \ min$,the concentration of $H_2O_2$ decreases from $0.5 \ M$ to $0.125 \ M$. For such decomposition,when the concentration of $H_2O_2$ reaches $0.05 \ M$,the rate of formation of $O_2$ will be:

For the reaction $2N_2O_5 \rightarrow 4NO_2 + O_2$,the rate constant is $3.0 \times 10^{-5} \ s^{-1}$. If the rate of reaction is $2.40 \times 10^{-5} \ mol \ L^{-1} \ s^{-1}$,calculate the concentration of $N_2O_5$ in $mol \ L^{-1}$.

For a first order reaction,the time required for completion of $90 \%$ reaction is '$x$' times the half life of the reaction. The value of '$x$' is $........$. (Given: $\ln 10 = 2.303$ and $\log 2 = 0.3010$)